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Topic: 0.999. (Read 124802 times) 

Icarus
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Re: 0.999.
« Reply #200 on: Mar 4^{th}, 2003, 6:24pm » 

0.999...=1 is no special case. Any "terminating" decimal (which is equivalent to an infinite decimal ending in repeating 0s) has a second representation ending in repeating 9s. The two decimal expressions are the same to the left of the last digit not zero in the terminating version. The last digit itself is one less for the repeating 9 version. For example, in addition the ones you listed, the following is also true: 3.44456 = 3.44455999... as is 9.999 = 9.998999... You may be relieved to know that terminating decimals are the only real numbers with more than one decimal expression, and they have only two each.

« Last Edit: Mar 4^{th}, 2003, 6:32pm by Icarus » 
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cho
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Re: 0.999.
« Reply #201 on: Mar 4^{th}, 2003, 7:48pm » 

Here's another proof that .999...=1 .999...=.9+.09+.009+.0009 which equals (1.1)+(.1.01)+(.01.001)... which equals 1+(.1+.1)+(.01+.01)+(.001+.001)... which equals 1+(.1.1)+(.01.01)+(.001.001)... which equals (1+.1)+(.1+.01)+(.01+.001)... which equals 1.000...1 which equals its reciprocal, 1+(1010)+(100100)+(10001000)... which equals (1+10)+(10+100)+(100+1000)... which equals 11+90+900+9000... which equals 1000000...1 ergo .999...=infinity


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schizo
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Re: 0.999.
« Reply #202 on: Mar 5^{th}, 2003, 5:38pm » 

x= 0.999... 10x= 9.999... 10x x= 9 9x= 9, x= 1 The numbers have to be finite in order to subtract since one must start from the smallest value of decimal place. In this case, the proof DOES assumes that there is a limit to 0.999...!


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S. Owen
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Re: 0.999.
« Reply #203 on: Mar 5^{th}, 2003, 5:54pm » 

Agreed... that solution assumes things that are seemingly more complex than the original problem (like what 10 * 0.999... means, what it means to subtract 0.999...) I don't find it very satisfying, since it doesn't seem to touch the "heart" of the issue. Heh, but at least it ends up at the right place.


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Icarus
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Re: 0.999.
« Reply #204 on: Mar 5^{th}, 2003, 8:40pm » 

An analysis of Cho's counter argument: .999...=.9+.09+.009+.0009 ... true which equals (1.1)+(.1.01)+(.01.001)... true which equals 1+(.1+.1)+(.01+.01)+(.001+.001)... This step is justifiable, but is not as obvious as Cho thinks. A convergent infinite series can only be rearranged if it is absolutely convergent. The basic sequence here, 1.1+.1.01+.01... is absolutely convergent, so it can be summed up in any order. So this step is true. which equals 1+(.1.1)+(.01.01)+(.001.001)... which equals (1+.1)+(.1+.01)+(.01+.001)... I have to assume Cho lost track of where he was going here. Both of these steps are justifiable, but the next step clearly follows from the line before these two, and it is NOT obvious that it is equal to this last line! which equals 1.000...1 Since we are clearly dealing with real numbers here, there is no such thing as "1.000...1" (dots refering to an infinite number of zeros). However, if this is changed to "1.000...", then this is an easy equivalent to the third expression. which equals its reciprocal, 1+(1010)+(100100)+(10001000)... true, once the previous expression is corrected which equals (1+10)+(10+100)+(100+1000)... Patently false. As I said earlier, you can rearrange how you sum an absolutely convergent series. But it a theorem that a conditionally convergent series (the series converges, but the series of absolute values does not converge) can be rearranged so as to converge to any limit, or to diverge to either oo or oo. The base series here 1+1010+100100... does not converge at all. You cannot rearrange how the sum is taken and expect it sum to the same value which equals 11+90+900+9000... True, in that [infty]=[infty] which equals 1000000...1 This is a number that fails to be real in two different ways, but Cho does give the correct value in the end: ergo .999...=infinity Of course, the equality does not hold because one of the intermediate steps is false. But notice that this misstep comes only after it is shown that 0.999... = 1. on Mar 5^{th}, 2003, 5:38pm, schizo wrote:The numbers have to be finite in order to subtract since one must start from the smallest value of decimal place. In this case, the proof DOES assumes that there is a limit to 0.999...! 
 If numbers had to be finite (in decimal places) to subtract, then subtraction could not be defined on anything but terminating decimals. You can start subtraction at the left and move right. Occasionally you have to move back left to borrow, but whenever a nonzero digit occurs in the number you are subtracting from, all future borrowings do not need to go any futher left than this digit, so the previous digits are all completely determined at this point. Thus as long as the number you are subtracting from does not terminate, you can subtract starting with the left and move right. If it does terminate, you can still subtract starting at the left: just write the number in its nonterminating form: X.dd...dd = X.dd...d(d1)999... In any case, this problem does not arise here, as there is no need to borrow to perform this subtraction. The existence of 0.999..., that 9.999...  0.999... = 9, and that 10 * 0.999... = 9.999... are all fairly fundamental properties of decimal notation, which I have a hard time believing anyone is unfamiliar with. It does not strike me that taking them for granted is inappropriate in a proof. After all, if every theorem had to be proven directly from the axioms, proofs would be exceedingly harder to follow than they already are! As for how satisfactory a proof it is, I suppose that depends on what you are looking for from your proof. This one (that schizo has reproduced) is short, sweet, and easily followed, which I find very satisfactory indeed, even if it doesn't give much insight as to why they are equal.

« Last Edit: Aug 19^{th}, 2003, 9:04pm by Icarus » 
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Kozo Morimoto
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Re: 0.999.
« Reply #205 on: Mar 5^{th}, 2003, 9:04pm » 

on Mar 4^{th}, 2003, 3:53pm, S. Owen wrote:I don't totally see the analogy with the jumprope, but I would note that the friend will always be a finite distance away, and is never "at infinity"... "infinity" is a place that the friend can go towards, but can never reach. 
 I never said the friend will ever be at infinity, just as how adding 9s to 0.9 will never ever become 0.999..., but "at infinity" by definition, its 0.999... So do I end up with 2 ropes of inifite length or do I still have 1 rope?


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David Ryan
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Re: 0.999.
« Reply #206 on: Mar 6^{th}, 2003, 3:20am » 

What is infiity times 2? All of my classes thus far have led me to believe it is infinity. If you have a rope doubled over at an infinite distance, it is still A rope, one. If the rope stretches "to infinity", then you, say, cut the middle, you would have 2 ropes, both of infinite length. That, at least, is the conclusion that I come to at 6:00 in the morning before school. Also, someone had mentioned don't get you guys started on different types of infinity; I was unaware that there were. Please...enlighten me. As a note, this site has made study hall one of my two most educational classes throughout the day. (what a nerd i am) //edited by Icarus to fix typo pointed out in the next post.

« Last Edit: Oct 25^{th}, 2003, 8:16am by Icarus » 
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David Ryan
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Re: 0.999.
« Reply #207 on: Mar 6^{th}, 2003, 3:23am » 

3rd line, 6th word = cut


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wowbagger
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Re: 0.999.
« Reply #208 on: Mar 6^{th}, 2003, 4:01am » 

on Mar 6^{th}, 2003, 3:20am, David Ryan wrote:Also, someone had mentioned don't get you guys started on different types of infinity; I was unaware that there were. Please...enlighten me. 
 For a brief intro, see Icarus's post a few pages up in this thread. Btw: I guess you meant the 7th word  unless you're interpreting your post as an array in C, in which case it would be the 2nd line Members can modify posts


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aero_guy
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Re: 0.999.
« Reply #209 on: Mar 6^{th}, 2003, 5:31am » 

Whew, that was less painful than I had thought, but then again I basically asked the same question in another thread just now, I guess I should review your link. Oh, and where the replacement goes depends on your browser width. OK guys, the horse is reduced to a puddle of goop, enough with the stick.


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S. Owen
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Re: 0.999.
« Reply #210 on: Mar 6^{th}, 2003, 6:59am » 

on Mar 5^{th}, 2003, 9:04pm, Kozo Morimoto wrote:So do I end up with 2 ropes of inifite length or do I still have 1 rope? 
 I see your analogy... I don't know if there is a "right" answer. I think that the only way you could define such a string of digits would be to take two strings starting with 1 (jumprope handles), with a bunch of 0's (jump rope), and say that their tails meet up at infinity. But, regardless of whether you can define a string like that, I don't think it means that it is a real decimal number. But you're right, neither is 0.999... In both cases you can still reasonably define a value for them by working out the limit  both have the value of 1, if anything. Interesting thought.


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Icarus
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Re: 0.999.
« Reply #211 on: Mar 6^{th}, 2003, 4:32pm » 

on Mar 6^{th}, 2003, 6:59am, S. Owen wrote:But, regardless of whether you can define a string like that, I don't think it means that it is a real decimal number. But you're right, neither is 0.999... In both cases you can still reasonably define a value for them by working out the limit  both have the value of 1, if anything. 
 You can define strings like that. For example, consider the set N2 = {1/n  n is a nonzero integer}. This set has as its lowest element 1, then 1/2, 1/3 ... infinitely many leading up to zero, then on the positive side, the same situation reversed, ending in 1. (Zero itself is not in the set.) A "sequence" from N2 into the digits {0,1,2,3,4,5,6,7,8,9} provides just such a string. That is, you index the digits not with natural numbers ("digit 1, digit 2, ..."), but with egyptian fractions ("digit 1, digit 1/2, ... ..., digit 1/2, digit 1"). This is somewhat clumsy, so instead call them "out" and "back" digits ("digit 1 out, digit 2 out, ... ..., digit 2 back, digit 1 back"). But the next thing you have to figure out is, if this is to represent a real number, what value does it have? Contrary to what S.Owen says 0.999... IS a real number. Its value is welldefined: 0.999... represents the unique real number L such that for any e>0, there is a natural number N such that for all n > N,  SUM_{i=1}^{n} 9*10^{i}  L  < e (in other words:  0.999...(n 9s)...9  L  < e ) That such an L exists and that it is unique have both been proven, for all sequences of digits, not just constant 9s. And it has been shown here many times that for 0.999..., that number L is 1. But this definition does not extend in any obvious way to the "outandback" style decimal Kozo describes. So there is no way to "work out the limit". The best you can do, staying in the Real numbers, is to define the value to be that of the digits on the "out" side, without regard to the "back" side. But if you are bothered by 2 decimal expressions for the same number, why would you want to add infinitely many more decimal expressions for it?


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Icarus
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Re: 0.999.
« Reply #212 on: Mar 17^{th}, 2003, 10:31pm » 

Since I'm feeling loquacious (i.e. bored) tonight, I thought it might be helpful to those who are interested, but have never seen it, to give the process by which the Real numbers are defined. There are actually several possible definitions, but this one is "standard". (More accurately, what I am doing here is the standard method of building a model for the Real numbers.) 1) (Naturals) We start with sets and cardinality. Two sets are called equipotent if there is a onetoone correspondance between all the elements of one and all the elements of the other. For example the sets {1,2,3} and {a,b,c} are equipotent since we can match up their elements: 1  a 2  b 3  c But we cannot match up the elements of {1,2,3,4} with {a,b,c}: 1  a 2  b 3  c 4  So these sets are NOT equipotent. Say A~B if the set A is equipotent to the set B. The three basic properties of equipotence are: for all sets A, B, C: (1) A~A (reflexivity) (2) If A~B, then B~A (symmetry) (3) if A~B and B~C, then A~C (transitivity). These 3 properties mean that ~ is an equivalence relation. (The next step actually produces paradoxes as I have it stated, if you look deep enough. But the methods for avoiding the paradoxes requires some highpowered math, so I am going to ignore these nasty facts instead.) For each set A, the set Card(A) = {B  B is equipotent to A} is called the cardinality of A. Objects of the form Card(A) for some A are called cardinals or cardinal numbers. Definition of finite and infinite: A set A is infinite if and only if there is a proper subset B of A such that A~B. A cardinal is called infinite if it is the cardinal of an infinite set. A cardinal (or set) is finite if it is not infinite. (For example, the set N of natural numbers  which I am in the middle of building  is infinite, since we have the onetoone correspondence x <> 2x between N and the set of even natural numbers.) We can define the following for cardinals (iff = "if and only if"): Order) Card(A) <= Card(B) iff A~C for some subset C of B. Sum) If A and B are disjoint (have no common elements), then Card(A) + Card(B) = Card(A U B). Product) For any sets A, B, the product set A x B = { (a,b)  a is in A and b is in B}. Define Card(A) * Card(B) = Card(A x B). Exponentiation) A subset f of A x B is called a map from A to B if for each a in A, there is exactly one b in B such that (a,b) is in f. (In this case, we define: f(a) = b.) Define the set B^{A} = {f  f is a map from A to B}. Define Card(B)^{Card(A)} = Card(B^{A}). Some other definitions: 0 = Card({}) (cardinal of the empty set) 1 = Card({0}) 2 = 1 + 1; 3 = 2 + 1; 4 = 3 + 1; ... 9 = 8 +1, 10 = 9 + 1. The set of all finite cardinals is called the Whole numbers, and is denoted by W. The set of Whole numbers greater than 0 is called the Natural numbers and is denoted by N. (It is also common to define the Natural numbers to include 0, and call the set without 0 "The Positive Naturals" N^{+}. The definitions I give here are more traditional, since the conception of Natural numbers predates the introduction of 0 into Western mathematics. They are also more common in my experience, as it is handy to have a quick reference set around that starts at 1 instead of 0.) 2) (Integers). Define the following relation on the set W^{2} = W x W: (a,b) ~ (c,d) iff a + d = b + c. It is easily verified that ~ is an equivalence relation, which allows us to define the notation [(a,b)] = {(c,d)  (c,d) ~ (a,b)}. ( [(a,b)] is called the equivalence class of (a,b).) The set Z = {[(a,b)]  a, b in W} is called the set of Integers. We can define Order) [(a,b)] < [(c,d)] iff a + d < b + c. Sum) [(a,b)] + [(c,d)] = [(a+c,b+d)] Difference) [(a,b)]  [(c,d)] = [(a+d,b+c)] Product) [(a,b)]*[(c,d)] = [(ac+bd,bc+ad)] The Whole numbers sit in the Integers by the mapping x > [(x,0)]. We suppress this map in the future and take the whole numbers (and the naturals) to be a subset of Z. We also note that if we define n = [(0,n)] as well, then we have Z = {..., 2, 1, 0, 1, 2, ...}, and we can write a  b for [(a,b)]. Accompanying these definitions are a whole slew of properties of <, +, , and * that everyone knows. 3) (Rationals) Let X = {(a,b)  a and b are integers and b != 0}. Define the following equivalence relation on X: (a,b) ~ (c, d) iff ad = bc. (Note that this would not be an equivalence if we allowed b or d to be 0). The equivalence classes are again [(a,b)] = {(c,d)  (c,d) ~ (a,b)}. The set of Rational Numbers is Q = {[(a,b)]  (a,b) is in X} The order relation and operators are defined by: Order) [(a,b)] < [(c,d)] iff (ad < bc with bd>0), or (ad > bc with bd < 0) Sum) [(a,b)] + [(c,d)] = [(ad+bc, bd)] Difference) [(a,b)]  [(c,d)] = [(adbc, bd)] Product) [(a,b)]*[(c,d)] = [(ac,bd)] Quotient) [(a,b)]/[(c,d)] = [(ad,bc)] (c != 0) We imbed Z into Q by the map x > [(x,1)], and from here on suppress the map. Then we can express [(a,b)] = a/b. 4) (Reals). First, a couple definitions: An upper bound for a set S is a number b such that for all x in S, x <= b. If S has an upper bound, then S is said to be bounded above. The supremum or least upper bound of a set S is a number B such that (i) B is an upper bound of S, and (ii) for every upper bound b of S, B <= b. The supremum of S is denoted by sup(S). If the supremum of S exists, it must be unique. But there are bounded sets S of rational numbers which do not have a supremum in Q. For example, the set S_{2}={x  x<=0 or (x>0 and x^{2} < 2)}. A subset S of Q is called a Dedekind Cut if (i) S is not empty, and is bounded above. (ii) For any x in S, if y is in Q and y < x, then y is in S (iii) S has no maximal element (i.e. for all x in S, there is a y in S such that x < y). Two examples: if r is a rational number, then the set { x  x<r } is a Dedekind cut, as is the set S_{2} defined above Define the set R of Real numbers to be R = {S  S is a Dedekind Cut of Q}. Order) If S, T are in R, then S <= T if S is a subset of T. Operations) S # T = {x  there is s in S and t in T with x <= s # t}, where # is any of the operations +, , *, or /. Q imbeds in R by the map x > {y  y < x}. Once again, we suppress the map and consider Q to be a subset of R. Example: A little work will show that S_{2}*S_{2} = 2, and S_{2} > 0, so S_{2} = sqrt(2). The Reals have one significant advantage over the Rationals: they possess a property called "completeness", whose easiest expression is the "supremum property": Theorem: (Supremum Property) If S is a nonempty subset of R, and S is bounded above, then S has a supremum. Proof: The Dedekind Cut {x  x in Q, and x < y for some y in S} is the supremum. (The "infimum property", is equivalent.) 5) (Decimal Notation). If you have made it through this, you will see that the Real numbers are defined entirely independent of the concept of decimal notation. This is as it should be, for decimals are only a way of denoting Real numbers. They are not what Real numbers are! For this reason I stopped the definitions of 0, 1, ... at 10 above. Now we define decimal notation: Definition: A (positive) decimal expression is a map D from the set <n> = { z  z is an integer, and z <= n} for some Whole number n into the set {0,1,2,3,4,5,6,7,8,9} of digits, with D(n) !=0 if n>0 (D(n) is the leading digit only). If D is a decimal expression, it is denoted (in the U.S.A.) by listing the digits in descending order of their indexes, with a "decimal point" between the digits for indexes 0 and 1: If d_{i} = D(i) for all i, then D is denoted by d_{n}d_{n1}...d_{0}.d_{1}... If there is an integer k <= 0 such that d_{i} = 0 for all i < k, then we usually drop all the trailing zeros from the notation: d_{n}d_{n1}...d_{0}.d_{1}...d_{k} and call it a terminating decimal. (Now you know! ) By definition, the decimal expression D represents the real number sup({ d_{n}*10^{n} + d_{n1}*10^{n1} + ... d_{k}*10^{k}  k in Z with k < n}). To complete the definition, D is the opposite of the number represented by D. Finally, let me make the following observations relating to comments posted in this forum. While it is possible to invent decimal notations such as 0.999...(infinite 9s)...9, they are not covered by this or any standard version of the definition of decimal expressions. Thus it is NOT enough to simply say "what about 1.000...01"? Decimal expressions are not numbers in and of themselves, they are only a way of denoting numbers. If you introduce new decimal expressions, YOU also have to explain what numbers they denote. On the other hand, any map from an initial segment of Z (i.e. a set <n> for some n) into the digits defines a decimal expression, which has a definite value as given above. This is standard for any accepted form of the definition of decimal expressions. In particular 0.999... and 1.000... are both valid decimal expressions and both represent real numbers which, as has been shown many times, are in fact the same. (Thanks to Wowbagger for pointing out some corrections!)

« Last Edit: Apr 14^{th}, 2003, 5:11pm by Icarus » 
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Chewdog
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Re: 0.999.
« Reply #213 on: Apr 14^{th}, 2003, 1:41am » 

In elementary we always learn that a/a = 1, correct? That is that any number divided by itself is always going to be once, since one nuber goes into itself exactly one time. Hence: The approximation that 3/3 = .999... itself is illogical and an invalid aproximation. Just because 1/3 is approximated in decimal form as .333... and 2/3 is approximated as .666... The correct assumption for 3/3 should be 1. This means that .999... itself is not even = to 3/3, however, and it is proof that .999<1


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BNC
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Re: 0.999.
« Reply #214 on: Apr 14^{th}, 2003, 4:49am » 

Here we go again......... Have you read the 9 pages of this thread before posting?


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aero_guy
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Re: 0.999.
« Reply #215 on: Apr 14^{th}, 2003, 5:33am » 

[Inigo]Let me 'splain, no there is too much, let me sum up...[/Inigo] You are using circular reasoning. You say that it is not equal because it is not equal. Your basic premise is wrong though, read the other messages.


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rmsgrey
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Re: 0.999.
« Reply #216 on: Apr 14^{th}, 2003, 12:24pm » 

Something to meditate upon for those who think that because 0.999...(finite)<1 0.999...(infinite)<1 1+1+1+...(finite)<oo 1+1+1+...(infinite)=? PS might not be too productive, but I felt that having waded through the thread I had to do something to justify my time... PPS To the best of my recollection, my maths course introduced the Reals as the closure under Cauchy limits of bounded monotone sequences of rationals. Of course, Icarus' definition is equivalent (I assume  at 5am my deductive processes are prone to undetected errors), but 'my' definition leads directly to all (nontransfinite) decimal values being reals. The pythagorean proof that root 2 is irrational suffices to show that the reals contain elements not found within the rationals. As I recall, Cantor's diagonalisation argument was used to prove different cardinality.


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Icarus
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Re: 0.999.
« Reply #217 on: Apr 14^{th}, 2003, 5:01pm » 

It's a long slog to the end of this thread, and those posters who just can't seem to say their piece in a few lines, but instead write epic responses are no help at all. There are several ways to complete the Rationals and obtain the Reals. All of them are equivalent. I chose Dedekind cuts because they are in some ways more elementary than the rest. They require a slightly lower level of mathematical sophistication to understand than Cauchy sequences do. In particular, you don't have to understand limits to understand them. And several posters are very confused when it comes to limits. (This is not uncommon. My experience is that less than 10% of all successful calculus students really have any idea what a limit is. The rest just manage to work things out by rote. And of those 10%, a majority have serious misconceptions.) There are also ways to complete the Rationals and get

« Last Edit: Apr 14^{th}, 2003, 5:02pm by Icarus » 
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Zeb Dahl
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Re: 0.999.
« Reply #218 on: May 2^{nd}, 2003, 1:13pm » 

on Jul 26^{th}, 2002, 5:16pm, Kozo Morimoto wrote:First off 1/3 is NOT 0.3333... that's just the decimal approximation. I don't understand the first solution. And 0.0000... is NOT 0, very close to zero, but not zero, so that equal sign in the middle is wrong. 1 is not equal to 0.999... I suppose you have to define what EQUAL is... I've heard that something with a probability of less than 1 out of 10^40 is mathematically classified as impossible, although it isn't 0. 



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Daniel J. Wells
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Re: 0.999.
« Reply #219 on: Jun 10^{th}, 2003, 2:43pm » 

I dont know where this discussion started, but I would like to end it. The question concerns the nature of the number 0.999... . The answer is simple. The number 0.999... is not a Real Number. If you want to work with the Hyperreal numbers then it is a nonstandard number with real part 1. Yes, this number 0.999... is less than 1, but it is also greater than every real number less than one. This sort of situation is not possible in teh Real numbers because they have the Archemidian property, that is. there is some multiple of 1 that is greater than any given real number. This also implies that for any possitive real number there is a smaller possitive number of the form 1/N where N is a multiple of 1. Suppose a number, Q (0.999...) is less than 1 but greater than every (other) real number less than 1. Clearly, 1Q is greater than 0, since Q<1. (1Q>11=0) Also since Q is greater than any real number less than 1, Q is less than any number greater than 1 and thus 1Q is less than any real greater than 0. But as I mentioned early the Arcemedian principle garuntees that there a natural number N such that 1/N<Q. Thus the supposition that Q exists is false. (Indeed, if you ignore the Completeness Property, and make this assumption, you obtain the Hyperreal Numbers.) [Please forgive any attrocious spelling errors I have most certainly made.]


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Daniel j Wells
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Re: 0.999.
« Reply #220 on: Jun 10^{th}, 2003, 2:51pm » 

Just in case you missed the point of my post: 0.99999999.... is not a Real Number (the kind you use in calculus, or algerbra for than matter.) any more than i (where i^2=1) is a Real Number.


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Sir Col
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Re: 0.999.
« Reply #221 on: Jun 10^{th}, 2003, 3:13pm » 

Are you suggesting that 1 is not a real number? s_{1}=1/2+1/4+1/8+1/16+... 2s_{1}=1+1/2+1/4+1/8+1/16+... Therefore 2s_{1}–s_{1}=s_{1}=1 In the same way, s_{2}=9/10+9/100+9/1000+... 10s_{2}=9+9/10+9/100+9/1000+... Therefore 10s_{2}–s_{2}=9s_{2}=9 and so s_{2}=1. In other words, as much as it is true to say that the infinite sum of 1/2+1/4+...=1, it is true that to say that 0.999...=9/10+9/100+9/1000+...=1.

« Last Edit: Jun 10^{th}, 2003, 3:14pm by Sir Col » 
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Icarus
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Re: 0.999.
« Reply #222 on: Jun 10^{th}, 2003, 3:44pm » 

I stand amazed at someone who has heard about hyperreals (and therefore presumably nonstandard analysis), the completeness property, and the archimedean property, but does not understand the meaning of 0.999... . As Sir Col indicates, the definition of 0.999... is [sum]_{n=1}^{[supinfty]} 9*10^{n}. How can you claim that this convergent series is not real?

« Last Edit: Aug 19^{th}, 2003, 6:30pm by Icarus » 
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"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



redPEPPER
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Re: 0.999.
« Reply #223 on: Jun 11^{th}, 2003, 3:33am » 

on Jun 10^{th}, 2003, 2:43pm, Daniel J. Wells wrote:Suppose a number, Q [...] is less than 1 but greater than every (other) real number less than 1. Clearly, 1Q is greater than 0, since Q<1. (1Q>11=0) Also since Q is greater than any real number less than 1, Q is less than any number greater than 1 and thus 1Q is less than any real greater than 0. But as I mentioned early the Arcemedian principle garuntees that there a natural number N such that 1/N<[1Q]. Thus the supposition that Q exists is false. 
 Congratulations, you just proved that 0.999... = 1 What? You don't see it? Well, you demonstrated that there cannot be a real number smaller than 1 but greater than any other real number smaller than 1. By notation, 0.999... is real, greater than any other real number smaller than 1, and certainly not greater than 1 itself. By your demonstration it cannot be smaller than 1. It must therefore be equal to 1 Your flaw was to assume from the very start that 0.999...=Q without proving it, specifically when you define Q<1. But 0.999...<1 is far from a given, as this whole thread shows.

« Last Edit: Jun 11^{th}, 2003, 3:36am by redPEPPER » 
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Sir Col
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Re: 0.999.
« Reply #224 on: Jun 11^{th}, 2003, 4:59am » 

One of the classic proofs I use with my classes when they're struggling with this concept is simple, yet powerful. Given any two numbers, a and b, the arithmetic mean, (a+b)/2, lies midway between a and b. If a=0.999... is the 'last' number before 1 and b=1. I would ask, what is the arithmetic mean of a and b? The only way to resolve this dilemma is for a and b to be equal. However, I must give some consideration to Daniel's argument. I have never heard of hyperreal numbers, so I will have to research them for myself before I am able to fully appreciate his points. Despite my ignorance I can accept that if we define 0.999... as a number outside of the real numbers, then it is not real – that is common sense; if the set of A (hyperreal numbers) are not part of the set B (real numbers) and c (0.999...) is contained in A (hyperreal numbers) then c (0.999...) is not part of B (real numbers). My issue would be whether such a set exists and then whether 0.999... belongs to this set. Obviously imaginary numbers are mutually exlusive to real numbers, but I thought that real and imaginary collectivey (the complex set) was complete; isn't that the fundamental theorem of arithmetic which Gauss proved? I do appreciate though that my proof above only holds if 0.999... is real in the first place and the other proofs that have been presented tease our intuition tantalisingly on the borders of the infinitesimal and the infinite – both concepts too small or too big by definition for us to comprehend. Is there any truth in this?

« Last Edit: Jun 11^{th}, 2003, 5:00am by Sir Col » 
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